Question

The sum of first three terms of a G.P. is 13/12 and their product is − 1. Find the G.P.

Answer 1

Let the first three numbers of the given G.P. be $\frac{a}{r}\mathit{,}\mathit{}a\mathrm{and}\mathit{}ar$.
∴ Product of the G.P. = −1
$\Rightarrow$ a³ = −1
$\Rightarrow$a = −1
Similarly, Sum of the G.P. = $\frac{13}{12}$
$\Rightarrow \frac{a}{r}+a+ar=\frac{13}{12}$
Substituting the value of a = −1
$$\begin{gathered} \frac{-1}{r}-1-r=\frac{13}{12} \\ \Rightarrow 12r^2+25r+12=0 \\ \Rightarrow 12r^2+16r+9r+12=0 \\ \Rightarrow 4r\left(3r+4\right)+3\left(3r+4\right)=0 \\ \Rightarrow \left(4r+3\right)\left(3r+4\right)=0 \\ \Rightarrow r=-\frac{3}{4},-\frac{4}{3} \end{gathered}$$
Hence, the G.P. for a = −1 and r = $-\frac{3}{4}$ is $\frac{4}{3},-1\mathrm{and}\frac{3}{4}$.

And, the G.P. for a = −1 and r = $-\frac{4}{3}$ is $\frac{3}{4},-1\mathrm{and}\frac{4}{3}$.